What is the pH of a solution of #10^(-8)# M sodium hydroxide ?

Right from the start, you should be able to predict that the pH of the solution will be higher than #7#. This is the case because the solution contains a strong base.

You could look at the concentration of the base and say that the pH will only be slightly higher than #7#, but it must come out to be higher than #7#.

The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to

#K_W = 10^(-14) -># water's ionization constant

In pure water, water undergoes self-ionization to form hydronium cations, #"H"_3"O"^(+)# and hydroxide anions, #"OH"^(-)#, as described by the following equilibrium reaction

#2"H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

By definition, the ionization constant is equal to

#K_W = ["H"_3"O"^(+)] * ["OH"^(-)]#

If you take #x# to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

#K_w = x * x = x^2#

This gets you

#x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)#

So, the self-ionization of water produces

#["H"_3"O"^(+)] = 10^(-7)"M "# and #" " ["OH"^(-)] = 10^(-7)"M"#

at room temperature. Now, your solution contains sodium hydroxide, #"NaOH"#, a strong base that dissociates completely to produce hydroxide anions in a #1:1# mole ratio.

Therefore, you're adding

#["OH"^(-)] = ["NaOH"] = 10^(-8)"M"#

to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.

If you take #y# to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

#" "2"H"_ 2"O"_ ((l)) rightleftharpoons" " "H"_ 3"O"_ ((aq))^(+) " "+" " " ""OH"_ ((aq))^(-)#

#color(purple)("I")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))#
#color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))#
#color(purple)("E")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)#

This time, the ionization constant will be equal to

#K_W = y * (10^(-8) + y)#

#K_w = y^2 + 10^(-8) * y#

This will get you

#y^2 + 10^(-8) * y - 10^(-14) = 0#

This quadratic equation will produce two solutions, one positive and one negative. Since #y# represents concentration, pick the positive one

#y = 9.51 * 10^(-8)#

This means that the equilibrium concentration of hydronium cations will be

#["H"_3"O"^(+)] = 9.51 * 10^(-8)"M"#

The pH of the solution is given by

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Plug in your value to find

#"pH" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))#

As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.

This is a tiny concentration for NaOH so I would expect the pH to be close to 7 or slightly greater.

Because the concentration is so small, we must take into account the ions that are produced from the auto - ionisation of water:

#sf(H_2OrightleftharpoonsH^++OH^-)#

For which #sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2))# at #sf(25^@C)#.

This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are #sf(10^(-7)color(white)(x)"mol/l")# respectively.

To get the total hydroxide concentration you might think you simply add #sf(10^(-7))# and #sf(10^(-8))#. This is not the case as these will not be equilibrium concentrations.

We need to apply Le Chatelier's Principle.

If we do a thought experiment we can imagine some pure water to which a tiny amount of sodium hydroxide is added such that the concentration of NaOH is #sf(10^-8color(white)(x)"mol/l"#.

We have now disturbed a system at equilibrium by adding extra #sf(OH^-)# ions. The system will act to oppose that change by reducing the number of #sf(OH^-)# ions and shifting to the left.

The reaction quotient #sf(Q)# is given by #sf([H^+][OH^-])#. This is now greater than #sf(K_w)# so the system responds by shifting the position of equilibrium to the left such that #sf(Q=K_w)#.

We can set up an ICE table using equilibrium concentrations to show this:

#" "sf(H_2O" "rightleftharpoons" "H^+" "+" "OH^-) #

#sf(color(red)(I)color(white)(xxxxxxxxx)" "10^-7" "(10^(-7)+10^(-8))#

#sf(color(red)(C)color(white)(xxxxxxx)" "-xcolor(white)(xxx)" "-x)#

#sf(color(red)(E)color(white)(xxxxxxxx)" "(10^(-7)-x)" "(1.1xx10^(-7)-x))#

This gives us:

#sf((10^-7-x)(1.1xx10^(-7)-x)=10^(-14))#

If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for #sf(x)#. I won't go into that here but it gives, ignoring the -ve root:

#sf(x=0.49xx10^(-8))#

We can now get the equilibrium concentration of #sf([H^+])#:

#sf([H^+]=10^(-7)-(0.49xx10^(-8))=9.51xx10^(-8)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log[9.51xx10^(-8)]=color(red)(7.02))#

This is an example of "The Common Ion Effect", hydroxide being the common ion in question.